I start with a red, a blue, a green and a yellow marble. I can
trade any of my marbles for three others, one of each colour. Can I
end up with exactly two marbles of each colour?

Where Can We Visit?

Stage: 3 Challenge Level:

Fantastic responses to this problem,
Benjamin and George discovered this:

We started with 42 and using the rule multiply by 2 and subtract by
5.
We discovered when any number with the factor of 2 is doubled
(multiplied by 2) it gives a units digit of 4, 8, 6, and back to 2
which we started with. Although you can get round this by
subtracting 5 it only then takes you to 7, 3 1 and 9 which with the
10 by 10 grid it becomes impossible to numbers with the factors or
multiples of 5 e.g. 5, 10 15...

Lydia and Jess explain their
findings:

We can not visit numbers which are multiples of 5 (those
ending in a 5/0), 99 or 97. We could not visit multiples of 5, as
starting from 42, multiplying by 2 or subtracting by 5 would not
get us to a multiple of 5. Multiplying 42 by 2 results in 84, you
cannot do this again as the result would be above the number 100;
however many times you minus 5 you wold only visit numbers ending
in 7 or 2. We could not visit 97 or 99 as they are not even meaning
we can not double any number to get there. both of these are closer
to 100 than 5 meaning we can not subtract 5 to there.

Lyman Shen explored other combinations
of multiplying and subtracting:

Solution for $\times$ 2 and
-5:

First I started with 1. By multiplying by two I can go to 2,
4, 8, 16, 32, 64. And I go down 64 from -5 all the way to 9 (4
would do the same thing again). And then I multiply by 2 again, and
get 9, 18, 36, 72. I can carry this on to get every number, except
I can't get any multiple of 5's. Also I can't make 97, and 99. I
can't get them because they are odd numbers and can't get it by -5.
I can get 98 from $\times$ 2 from 49, which could be get from -5
form numbers end with 9 and 4.

Solution for $\times$ 3 and
-5:

By working similarly to before, I can find all the numbers
except for the multiple of 5's because no numbers can make it to 5,
10, unless I start with a multiple of 5. I also can't get 92, 97,
98. I can't get them because they aren't multiple of 3's and can't
get it from -5.

Solution for $\times$ 4 and
-5:

This is time the numbers that can be reached depend on the
numbers started with. When the starting number ends with 1, 4, 6,
9, there are no way to get any number that ends differently ($1
\times 4 = 4, 4 \times 4 = 16, 6 \times 4 = 24, 9 \times 4 = 36$).
So we can reach everything ending in 1, 4, 6, 9 except the large
numbers 94, 99. I can't get those number because it is not
divisible from 4 and I can't get them from -5, which I can do to
96.

When the starting number ends with 2, 3, 7, 8, there are no
way to get any number that ends differently ( $2\times 4 = 8,
3\times 4 = 12, 7\times 4 = 28, 8\times 4 = 32$). So I can reach
anything ending in these digits, apart from 97. I can't get it
because it isn't divisible from 4 and can't get it from -5.
When the start number ends with 5 or 0, there are no way to get any
number that ends in anything other than 0 and 5.

Solution for $\times$ 5 and
-5:

This reaches every multiple of 5, and the number you start
with.

The reason is $5\times x = 5x$, which have to end with 0 or 5.
No more exceptions after that.

Solution for $\times$ 6 and
-5:

This is another one that is dependant on the starting
number.

If it ends with 1 or 6, you can't get any numbers that don't
end with 1 or 6 ($1 \times 6 = 6, 6 \times 6 = 36$).

If it ends with 0 or 5, you can't get any numbers that don't
ends with 0 or 5 ($0 \times 6 = 0, 5 \times 6 = 30$). The ones you
can't reach 95 and 100 ($15 \times 6 = 90$).

If the stating number ends with 3 or 8. $3 \times 6 = 18$, $8
\times 6 = 48$ so again you can reach all numbers ending in 3 or 8,
but missing 98, 93, 88, 83 as the highest that can be reached is
$13 \times 6 = 78$

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